Quotients of C star algebras by closed two-sided ideals

For every $a\in A$ one has $\mathrm{\parallel }a+J\mathrm{\parallel }={\lim }_{\lambda }\mathrm{\parallel }(1-u_{\lambda })a\mathrm{\parallel }={\lim }_{\lambda }\mathrm{\parallel }a(1-u_{\lambda })\mathrm{\parallel }$, and the induced involution is isometric: $\mathrm{\parallel }{a}^{\ast }+J\mathrm{\parallel }=\mathrm{\parallel }a+J\mathrm{\parallel }$. Indeed, $(1-u_{\lambda })a=a-u_{\lambda }a$ with $u_{\lambda }a\in J$ gives $\mathrm{\parallel }(1-u_{\lambda })a\mathrm{\parallel }\ge \mathrm{\parallel }a+J\mathrm{\parallel }$; conversely for $j\in J$ one has $(1-u_{\lambda })a=(1-u_{\lambda })(a+j)-(1-u_{\lambda })j$, so $\mathrm{\parallel }(1-u_{\lambda })a\mathrm{\parallel }\le \mathrm{\parallel }a+j\mathrm{\parallel }+\mathrm{\parallel }(1-u_{\lambda })j\mathrm{\parallel }$ and for fixed $j$ the last term tends to $0$ by [F1], whence ${\mathrm{lim\hspace{0.17em}sup}}_{\lambda }\mathrm{\parallel }(1-u_{\lambda })a\mathrm{\parallel }\le \mathrm{\parallel }a+j\mathrm{\parallel }$ and, infimizing over $j$, ${\mathrm{lim\hspace{0.17em}sup}}_{\lambda }\mathrm{\parallel }(1-u_{\lambda })a\mathrm{\parallel }\le \mathrm{\parallel }a+J\mathrm{\parallel }$; the same computation on the right gives the second identity. Taking adjoints and using $u_{\lambda }^{\ast }=u_{\lambda }$ yields $\mathrm{\parallel }{a}^{\ast }+J\mathrm{\parallel }=\lim \mathrm{\parallel }(1-u_{\lambda })a^{\ast }\mathrm{\parallel }=\lim \mathrm{\parallel }a(1-u_{\lambda })\mathrm{\parallel }=\mathrm{\parallel }a+J\mathrm{\parallel }$.

Every injective star-homomorphism $\phi :E\to F$ is isometric. It is contractive by [F2]. If $h=h^{\ast }\in E$ had $r:=\mathrm{\parallel }\phi (h)\mathrm{\parallel }<\mathrm{\parallel }h\mathrm{\parallel }$, choose ${\lambda }_0\in \sigma (h)$ with $\mathrm{\mid }{\lambda }_0\mathrm{\mid }=\mathrm{\parallel }h\mathrm{\parallel }$ and a continuous $f$ on $[-\mathrm{\parallel }h\mathrm{\parallel },\mathrm{\parallel }h\mathrm{\parallel }]$ vanishing on $[-r,r]$ but not at ${\lambda }_0$. In particular $f(0)=0$. By [F5] choose polynomials $p_n$ with zero constant term converging uniformly to $f$. Then $p_n(h)\to f(h)\in E$ and $p_n(\phi (h))\to f(\phi (h))$ by [F4], while multiplicativity and linearity give $\phi (p_n(h))=p_n(\phi (h))$ without any unitality assumption. Boundedness of $\phi$ gives $\phi (f(h))=f(\phi (h))=0$, although $f(h)\mathrm{\ne }0$ by [F4], contradicting injectivity. Hence $\mathrm{\parallel }\phi (h)\mathrm{\parallel }=\mathrm{\parallel }h\mathrm{\parallel }$ for self-adjoint $h$, and the C*-identity gives $\mathrm{\parallel }\phi (a){\mathrm{\parallel }}^2=\mathrm{\parallel }\phi (a^{\ast }a)\mathrm{\parallel }=\mathrm{\parallel }{a}^{\ast }a\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^2$ for arbitrary $a$.

The quotient satisfies the C*-identity: $\mathrm{\parallel }a+J{\mathrm{\parallel }}^2=\mathrm{\parallel }{a}^{\ast }a+J\mathrm{\parallel }$ for every $a$. Indeed, by step 1.1 twice, $\mathrm{\parallel }a+J{\mathrm{\parallel }}^2={\lim }_{\lambda }\mathrm{\parallel }a(1-u_{\lambda }){\mathrm{\parallel }}^2$ and $\mathrm{\parallel }a(1-u_{\lambda }){\mathrm{\parallel }}^2=\mathrm{\parallel }(1-u_{\lambda })a^{\ast }a(1-u_{\lambda })\mathrm{\parallel }$; writing $(1-u_{\lambda })a^{\ast }a(1-u_{\lambda })=(1-u_{\lambda })(a^{\ast }a+j)(1-u_{\lambda })-(1-u_{\lambda })j(1-u_{\lambda })$ for $j\in J$ and using $\mathrm{\parallel }1-u_{\lambda }\mathrm{\parallel }\le 1$ gives ${\mathrm{lim\hspace{0.17em}sup}}_{\lambda }\mathrm{\parallel }(1-u_{\lambda })a^{\ast }a(1-u_{\lambda })\mathrm{\parallel }\le \mathrm{\parallel }{a}^{\ast }a+j\mathrm{\parallel }+{\lim }_{\lambda }\mathrm{\parallel }(1-u_{\lambda })j(1-u_{\lambda })\mathrm{\parallel }=\mathrm{\parallel }{a}^{\ast }a+j\mathrm{\parallel }$, so $\mathrm{\parallel }a+J{\mathrm{\parallel }}^2\le \mathrm{\parallel }{a}^{\ast }a+J\mathrm{\parallel }$ after infimizing over $j$. The reverse inequality is submultiplicativity in the quotient Banach algebra of [F3] together with the isometric involution of step 1.1: $\mathrm{\parallel }{a}^{\ast }a+J\mathrm{\parallel }\le \mathrm{\parallel }{a}^{\ast }+J\mathrm{\parallel }\mathrm{\parallel }a+J\mathrm{\parallel }=\mathrm{\parallel }a+J{\mathrm{\parallel }}^2$.

$A\mathrm{/}J$ is a C*-algebra: it is a Banach algebra by [F3], its involution is isometric by step 1.1, and it satisfies the C*-identity by step 2.1. The quotient map $q$ is contractive; if $A\mathrm{/}J\mathrm{\ne }0$ choose a nonzero coset $a+J$ and representatives $a+j_n$ with $\mathrm{\parallel }a+j_n\mathrm{\parallel }\to \mathrm{\parallel }a+J\mathrm{\parallel }$; the unit vectors $(a+j_n)\mathrm{/}\mathrm{\parallel }a+j_n\mathrm{\parallel }$ have images of norm $\mathrm{\parallel }a+J\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }a+j_n\mathrm{\parallel }\to 1$, so $\mathrm{\parallel }q\mathrm{\parallel }=1$, while $q=0$ and $\mathrm{\parallel }q\mathrm{\parallel }=0$ when $A\mathrm{/}J=0$. A star-homomorphism $\psi :A\to B$ with $J\subseteq \ker \psi$ kills $J$, hence induces a well-defined star-homomorphism $\bar{\psi }:A\mathrm{/}J\to B$ with $\psi =\bar{\psi }\circ q$, its boundedness follows from $\mathrm{\parallel }\psi (a)\mathrm{\parallel }=\mathrm{\parallel }\psi (a+j)\mathrm{\parallel }\le \mathrm{\parallel }\psi \mathrm{\parallel }\mathrm{\parallel }a+j\mathrm{\parallel }$ for every $j\in J$ by taking the infimum, and it is unique because $q$ is surjective.

Every star-homomorphism $\psi :A\to B$ between C*-algebras has closed image: factor $\psi$ through $A\mathrm{/}\ker \psi$ by step 3.1, obtaining an injective star-homomorphism $\bar{\psi }:A\mathrm{/}\ker \psi \to B$, which is isometric by step 1.2; the image $\bar{\psi }(A\mathrm{/}\ker \psi )$ is complete as the isometric image of a complete space, hence closed in $B$, and it equals $\psi (A)$.

The Axiom of Choice is inherited from the approximate-unit and calculus suppliers of [F1]–[F4]; the quotient norm and factorization arguments add no further choice (The Axiom of Choice).